Calc/Trig Help

Simplify y = f(x) = arccos (cos^4x - sin^4x) ; just one out of infinity many answers is good.

Clearer problem.

Use Wolfram Alpha.

Simplify y = f(x) = arccos (cos^4x - sin^4x) ; just one out of infinity many answers is good.

Can you distinguish ( ) better on the 4x? It seems that you can use the sin^2 + xos^2 = 1 trick to simplify and see what happens from there

Oops, sorry, didn't clarify it good enough.

The 4 is the exponent, and the x is a regular old number....so like .

I don't know the Wolfram Alpha. I'm also not quite sure you can pull the sin^x + cos^x = 1, as you can't pull out a ^2, can you?

Edit:

how2Wolfram Alpha

I would not call that "simplified".

Edit: Turns out you can distribute the powers, never mind.

Wait, what? No you can't. You can use difference of squares, though.

cos^4 - sin^4 = (cos^2)^2 - (sin^2)^2 = (cos^2 + sin^2)(cos^2 - sin^2) = 1(cos^2 - sin^2) = cos 2x.

arccos cos 2x = 2x.

Ah, I meant difference of squares.

Dam Rust did it before me. Yeah it turns out y = 2x

Wait, what? No you can't. You can use difference of squares, though.

cos^4 - sin^4 = (cos^2)^2 - (sin^2)^2 = (cos^2 + sin^2)(cos^2 - sin^2) = 1(cos^2 - sin^2) = cos 2x.

arccos cos 2x = 2x.

Just have to be a bit careful with arccos ( cos 2x). Arccos maxes out at pi, where as 2x doesn't. So if you plug in pi to both sides of the equations and work them out the long way, you wont get the same answer.

Notedly you get 0 = 2*PI

To add to what shouri said, if you graph the function f(x)=arccos(cos x), you end up with a wave, not a line. Subsequently, when you integrate it, you get another wave, not a parabola.

Another question. Expect more questions as I'm redoing a test and trying to figure out the ways to do the questions.

Say 4* (sin4x) / (sin2x) and I'm trying to find the limit as x->0.

Can I cancel out a sin2x from the numerator and denominator? If not, how do I find the limit?

If you're allowed to do derivatives, take the derivative of the top and bottom seperately then plug in.

If not... then F3...Sec.

Can't do derivatives, or L'Hopital's rule for that matter. I basically can only do substitution, factoring, sinc functions, or trig identities.

Yeah, L'Hopital makes it nice:
4 sin 4x / sin 2x => 16 cos 4x / 2 cos 2x => 16 cos 0 / 2 cos 0 = 16/2 = 8.

But there's also another way, using the sine double angle identity:
4 sin 4x / sin 2x => 4 sin 2(2x) / sin 2x => 4 (2 sin 2x cos 2x) / sin 2x => 8 cos 2x => 8 cos 0 = 8.

Great, three problems out of ten left that I don't get!

1. I need to find the limit when x => 0, without using sinc functions (I need 2 ways to solve it). I can't use L'Hopital or derivatives. Figured out.

2. Same as above: find the limit when x =>0. I can use sinc functions for this one.

3. I'll try to solve the third one for myself. Figured out.

Edit: Solved all of them!

I really don't feel like making a new thread.

"Known (3,2) is a point on the curve 2x^2-^3-10=0. Find the value of y when x = 3.04; use differentials in computing the approximation."

Not algebra. Differentials.

I can't read the formula... what is "-^3"?


Anyway, you probably want to basically add dx * d/dx (first approximation), then dx^2/2 * d^2/dx^2 (second approximation) and so forth until they become zero.


That is, if you know the value at a point (x = 3, y = 2) and you know the slope (d/dx = 3) then at x = 3.04, you have 0.04*3 + 2 = 2.12 as an approximation of (3.04, 2.12)


It's called the Taylor expansion, f(x + t) = f(x) + t/1!*f'(x) + t^2/2!*f''(x) + t^3/3!*f'''(x) + ...